Disassembly of a `int fn(); fn();` vs `int fn(void); fn();`

Asked Jun 20 '17 at 15:54

Active Jun 20 '17 at 16:08

Viewed 47 times

On 86_64:

void x()
{
    void y(void);
    y();
}

disassembles to

x:
        jmp     y

whereas

void x()
{
    void y();
    y();
}

disassembles to

x:
        xorl    %eax, %eax
        jmp     y

IOW, why does eax need to be zeroed before making a prototypeless call?

edited Jun 20 '17 at 16:08

asked Jun 20 '17 at 15:54

Petr Skocik

What level of optimization did you specify? – Jonathan Leffler Jun 20 '17 at 15:59
2

Because without a prototype it may be a varargs function, and for those the sysv ABI requires `al` to contain number of SSE registers used (zero here). – Jester Jun 20 '17 at 15:59
@JonathanLeffler 3 of course :) https://godbolt.org/g/gseo2H – Petr Skocik Jun 20 '17 at 16:00
@Jester: that sounds plausible — but it's defensive. You can't legitimately call a varargs function without a proper prototype ending with ellipsis `int y(…, ...)` in scope, so `y` can't be a varargs. Officially. Funnier things have been known. – Jonathan Leffler Jun 20 '17 at 16:01
@Jester Thanks. I knew it was here somewhere. :) – Petr Skocik Jun 20 '17 at 16:05
I'll observe, slightly tongue-in-cheek, that I'd be happier if the functions were declared to return `void` since they don't return a value. – Jonathan Leffler Jun 20 '17 at 16:07
1

@Jester: Even if it is a varargs function, the compiler is not required to account for that. Calling varargs function without seeing a *prototype* has always been *undefined behavior* in standard C. C compiler is not required to accomodate for the fact that `fun` can potentially be varargs function. If some compiler does that, it is a peculiarity of specific implementation, not explainable in any way by rules of standard C language. – AnT stands with Russia Jun 20 '17 at 16:20

0 Answers0